Answer:
x−2x4+2x3−7x2−8x+12=x3+4x2+x−6
The rational root theorem suggests that other possible roots may be -6, 6, -3, 3, -2, 2, -1, and 1. It turns out that x=-2x=−2 is a root, since (-2)^3+4(-2)^2+(-2)-6=0(−2)3+4(−2)2+(−2)−6=0 , so x+2x+2 is also a factor and we have
\dfrac{x^4+2x^3-7x^2-8x+12}{(x-2)(x+2)}=x^2+2x-3(x−2)(x+2)x4+2x3−7x2−8x+12=x2+2x−3
Finally, we can factorize the remaining quotient easily:
x^2+2x-3=(x+3)(x-1)x2+2x−3=(x+3)(x−1)
so the other factors are x+2x+2 , x+3x+3 , and x-1x−1 .
Answer:
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given
Required
Find the standard matrix
The standard matrix (A) is given by
Where
![T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=T%28x%29%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
becomes
![Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=Ax%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
The x on both sides cancel out; and, we're left with:
![A = [T(e_1)\ T(e_2)\ T(e_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D)
Recall that:
In matrix:
is represented as: ![\left[\begin{array}{c}a\\b\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D)
So:
![T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]](https://tex.z-dn.net/?f=T%28e_1%29%20%3D%20%281%2C2%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
![T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_2%29%20%3D%20%28-4%2C6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-4%5C%5C6%5Cend%7Barray%7D%5Cright%5D)
![T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_3%29%20%3D%20%282%2C-6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-6%5Cend%7Barray%7D%5Cright%5D)
Substitute the above expressions in
Hence, the standard of the matrix A is:
Step-by-step explanation:
- ATTACHED IS THE SOLUTION!
Probability =
(number or amount of possible successes)(total number or amount of possible outcomes)
Area of any circle is (pi) x (radius squared)
Area of the whole board = (pi) x (13 squared) = 169 pi square inches
Area of the shaded region = (pi) x (5 squared) = 25 pi square inches
Probability = (success area) / (total possible area) = 25pi/169pi = <u>0.15 </u>(rounded)
Notice that if you do it my way ... keep 'pi' in the fraction ... you don't even
need to know what the value of 'pi' is, because there's one in the top and
bottom of the fraction, and they cancel.
